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The upper 3/4th portion of a vertical pole subtends an angle tan^{-1}(3/5)   at a point in the horizontal plane through its foot and at a distance 40 m from the foot. A possible height of the vertical pole is

  • Option 1)

    40 m

  • Option 2)

    60 m

  • Option 3)

    80 m

  • Option 4)

    20 m


Answers (1)


As we learnt in 

Height and Distances -

The height or length of an object or the distance between two distant objects can be determined with the help of trigonometric ratios.


 \Theta =\tan ^{-1}\frac{3}{5}

Now \tan\Theta =\tan (\angle ACB-\angle DCB)

= \frac{\tan(\angle ACB-\angle DCB)}{1+\tan (\angle ACB)\tan \angle DCB}\, \, \, \, \,......(i)

In \Delta ACB, \tan \angle ACB=\frac{h}{BC}                -----(ii)

Similarly, \Delta DCB, \tan \angle DCB=\frac{h}{4BC}


using (i) & (ii)

\tan \Theta =\frac{3}{5} = \frac{\frac{h}{BC}-\frac{h}{4BC}}{1+\frac{h^{2}}{4(BC)^{2}}}=\frac{BC[3h]}{4BC^{2}+h^{2}}

\Rightarrow \frac{3}{5}=\frac{120h}{6400+h^{2}}

\Rightarrow 3h^{2}-600h+19200=0

solving, h=40\ or\ h=160


Option 1)

40 m

This is correct option

Option 2)

60 m

This is incorrect option

Option 3)

80 m

This is incorrect option

Option 4)

20 m

This is incorrect option

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